A right; that's what I thought it meant. Thanks for clarifying. I was unsure what we were trying to connect Gauss to via a path - I understand fully now. Thanks
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Anon888Jun 13 '11 at 22:46

4

@Qiaochu: why don't you provide what you wrote in your comment as an "official" answer, so the question has a helpful (as your comment is/was) answer. Just a thought. (I don't mean to be telling you what to do!)
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amWhyJun 13 '11 at 23:22

1 Answer
1

"Interestingly, Gauß is the only vertex that needs to be connected by paths with more than one edge."

As noted by Qiaochu in the comments, the graph is nonplanar because it contains a subdivision of $K_{3,3}$. However, it does not contain $K_{3,3}$ as a subgraph (only as a topological minor). The diagram shows the $K_{3,3}$ subdivision and that the Gauss node is connected to Weierstrass by a path of length $2$ and connected to Kummer by a path of length $3$.